Balanced Number (思维数位dp)

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job 
to calculate the number of balanced numbers in a given range [x, y].

Input

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10 18).

Output

For each case, print the number of balanced numbers in the range [x, y] in a line.

Sample Input

2
0 9
7604 24324

Sample Output

10
897

数位dp难在dp数组所储存的状态

此题枚举配合dp枚举平衡杠杆的位置

dp数组三维分别代表数位 位置 当前加和

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
long long dp[20][20][2000];
int a[20];
long long dfs(int lim,int pos,int k,int sum,int now)
{
    if(pos==0) return sum==0;
    long long ans=0;
    if(!lim&&dp[pos][k][sum]!=-1) return dp[pos][k][sum];
    int s=lim?a[pos]:9;
    for(int i=0;i<=s;i++)
    {
        ans+=dfs(lim&&i==s,pos-1,k,sum+now*i,now-1);
    }
    if(!lim) dp[pos][k][sum]=ans;
    return ans;
}
long long solve(long long x)
{
    if(x==-1) return 0;
    int cnt=0;
    while(x!=0)
    {
        cnt++;
        a[cnt]=x%10;
        x/=10;
    }
    long long ans=1;
    for(int i=1;i<=cnt;i++)
    {
        ans+=dfs(1,cnt,i,0,cnt-i)-1;
    }
    return ans;
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long temp1,temp2;
        scanf("%lld%lld",&temp1,&temp2);
        cout<<solve(temp2)-solve(temp1-1)<<endl;
    }
}